Question

can u clearly state the answer please

You may need to use the appropriate technology to answer this question. A Pew Research study conducted in 2017 found that approximately 75% of Americans believe that robots and computers might one day do many of the jobs currently done by people.T Suppose we have the following data collected from nurses, tax auditors, and fast food workers in which a higher score means the person feels his or her job is more likely to be automated. Nurse Tax Auditor Fast Food Worker 4 45 4 5 7 337 4 5 6 5 26 automated for the three professions (a) Use 0.05 to test for differences in the belief that a person's job state the run and siternative hypotheses. Ho Tot all the population means are equal e xaudio Ft-food worker otse Test e r exortowe ** S9

Answer 1

This is ANOVA - Single Factor. We first need to find the $SST_{T}$ and $SST_{B}$. They are 'Total variance' and 'Variance in between the treatments' respectively.

We know

$SST_{r}=SST_{T}-SST_{B}$

$SST_{T}$ = $Sum \:of\: individual \:squares - \frac{Grand\:total^{2}}{n}$

Grand total = 110 Sum of individual squares = 566

$SST_{T}$ = 61.833

$SST_{B}$ = $\sum (\frac{square\:of\:total\:of\:individual\:treatment\:score}{no. \:of\:obs\:in\;that\;treatment})-\frac{Grand\:total^{2}}{n}$

Note: We have to separately sum $\frac{Square\;of\;total\:of\:treatment}{obs}$ for all the treatment.

$\sum (\frac{square\:of\:total\:of\:individual\:treatment\:score}{no. \:of\:obs\:in\;that\;treatment})$ = $\frac{30^{2}}{8}+\frac{32^{2}}{8}+\frac{48^{2}}{8}$

= 528.5

$SST_{B}$ = 24.333

$SST_{r}=$ 37.5

We can also use excel data analysis function

Steps:

1. Input data
2. Click on 'Data' on toolbar and then click on data analysis
3. Select 'Anova- Single Factor'
4. Input your range and other info accordingly and you will get your analysis.

ANOVA
Source of Variation	SS	df	MS SS/df	F
Between Groups	24.333	2  ( k -1)	12.167	6.813
Within Groups	37.5	21  (n-k)	1.786
Total	61.833	23 (n-1)

a.

This is a test for difference of means. So the null hypothesis will be

Null hypo: $\mu$ nurse = $\mu$ TA = $\mu$ FFW

Alternative hypo: Not all the populations are equal.

b.

F - stat = MSB / MSR

= 12.167 / 1.786

F -stat = 6.813

c. p - value is the probability of null hypothesis being true

p - value = P($F_{dfb,dfr}$ > Test Stat )

= P( $F_{2,21}$ > 6.813)

p-value = 0.005 ...........using f-dist tables with df = 2 and df = 21

d. Since  p-value < 0.05

We reject the null hypothesis and there is sufficient evidence for a difference.

Option 3

e.

LSD = $\sqrt{MSR(1/n1+1/n2)}$

The n1 and n2 depend on the sample of the treatments.

LSD = $\sqrt{1.786(1/8+1/8)}$ ...........since all the sample sizes are same.

LSD = 0.668

|Mean nurse - Mean TA| = |30/8 - 32/8|

|Mean nurse - Mean TA| = 0.25

Since the mean difference < LSD

There is not significant difference between the means for nurses and tax auditors.